Problem: Solve for $x$ : $2x^2 + 16x + 32 = 0$
Dividing both sides by $2$ gives: $ x^2 + {8}x + {16} = 0 $ The coefficient on the $x$ term is $8$ and the constant term is $16$ , so we need to find two numbers that add up to $8$ and multiply to $16$ The number $4$ used twice satisfies both conditions: $ {4} + {4} = {8} $ $ {4} \times {4} = {16} $ So $(x + {4})^2 = 0$ $x + 4 = 0$ Thus, $x = -4$ is the solution.